Computing $\lim_{n\rightarrow\infty}(1-\frac{x}{n})^{-n}$ My question is how to argue the following statement
$$\lim_{n\rightarrow\infty}\left(1-\frac{x}{n}\right)^{-n} = e^{x}.$$
My solution is using the binomial series of $\left(1-\frac{x}{n}\right)^{-n}$ followed by taking the limit and finally converting back into $e^{-x}$.
I'm wondering if there is a more straightforward way to prove this, saying only limit computations.
And my definition of exponential function is given as following
$$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n} = e^{x}$$
 A: An idea using what you know and a little arithmetic of limits: 
$$\lim_{n\to\infty}\left(1-\frac{x}{n}\right)^{-n}=\lim_{n\to\infty}\left[\left(1+\frac{(-x)}{n}\right)^n\right]^{-1}=\left(e^{-x}\right)^{-1}=e^x$$
A: If we already have the basic properties of the exponential function, we can instead calculate
$$\lim_{t\to 0^+} (1-tx)^{-1/t}.$$
To compute this, note that the logarithm of our expression is 
$$-\frac{\log(1-xt)}{t}.$$
One round of L'Hospital's Rule will find the limit of this. The result is $x$, so the limit of your expression is $e^x$.
Remark: It is $e^x$, not $e^{-x}$. 
If we have defined $e^w$ as the limit of $(1+w/n)^n$, then essentially no calculation is needed, since $\left(1-\frac{x}{n}\right)^{-n}=\frac{1}{\left(1-\frac{x}{n}\right)^{n}}$. 
A: While other answers have assumed $e^{x} e^{-x} =1$ this does require an effort to prove if we use the definition of $e^{x} $ given in the question and is in fact the key to the solution. We can easily note that if $|x|<n$ then by Bernoulli's inequality we have $$1-\frac{x^{2}}{n}\leq \left(1-\frac{x^{2}}{n^{2}}\right)^{n}\leq 1$$ and hence we have via Squeeze theorem $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}\left(1-\frac{x}{n}\right)^{n}=1$$ and this means that $$\lim_{n\to\infty} \left(1-\frac{x}{n}\right)^{-n}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}$$
